Question

Calculus AB

Answer 1

Assuming you mean
$\lim_{x \to 4} \frac{\sqrt{x+5}-3}{4-x}$

that means as x approaches 4

if we sub 0 for x we get
0/0
and intermitent form
use l'hopital's rule

so
take the derivitive of the top and bottom seperatly
l'hopitals rule is something like
if $\lim_{x \to n} \frac{f(x)}{g(x)}$ results in 0/0 or -∞/∞ or∞/∞ then keep doing it until f(n)/g(n) gives a form that isn't intermitent

so

take derivitive of top and bottom
$\dfrac{\frac{1}{2\sqrt{x+5}}}{-1}$
now, if we subsitute 4 for x we get
$\dfrac{\frac{1}{2\sqrt{4+5}}}{-1}$=

$\dfrac{\frac{1}{2\sqrt{9}}}{-1}$=

$\dfrac{\frac{1}{2(3)}}{-1}$=

$\dfrac{\frac{1}{6}}{-1}$=

$\dfrac{1}{-6}=\dfrac{-1}{6}$